LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Sequence basics in an arbitrary ordered field: limits are unique, limits preserve non-strict inequalities, convergent sequences are Cauchy, Cauchy sequences are bounded, and a Cauchy sequence with a convergent subsequence converges

Statement

Let $F$ be an ordered field (Ordered field) and let $(x_{k})$ , $(y_{k})$ be sequences in $F$ , with convergence in $F$ , Cauchyness in $F$ , boundedness and subsequences as in Sequences, convergence, Cauchyness, monotonicity, boundedness and closed intervals in an arbitrary ordered field. Then:

Limits are unique. If $x_{k} \to L$ and $x_{k} \to L^{'}$ in $F$ , then $L = L^{'}$ . A convergent sequence therefore has exactly one limit in $F$ and the notation $\lim_{k} x_{k}$ denotes it unambiguously. This is the licence under which the remaining clauses are written as equations between limits, and it is not new here: Sequences, convergence, Cauchyness, monotonicity, boundedness and closed intervals in an arbitrary ordered field already establishes it, in an arbitrary ordered field and with no completeness or Archimedean hypothesis. It is restated as clause 1 so that this lemma is self-contained as the citation target of the whole abstract chain on this page.
Limits preserve non-strict inequalities. If $(x_{k})$ and $(y_{k})$ both converge in $F$ and $x_{k} \leq y_{k}$ for every $k$ , then

$\lim_{k} x_{k} \leq \lim_{k} y_{k} .$
Convergent implies Cauchy. If $(x_{k})$ converges in $F$ , it is Cauchy in $F$ .
Cauchy implies bounded. If $(x_{k})$ is Cauchy in $F$ , it is bounded.
A Cauchy sequence with a convergent subsequence converges. If $(x_{k})$ is Cauchy in $F$ and some subsequence $(x_{n_{j}})$ converges in $F$ , then $(x_{k})$ converges in $F$ as well, and

$\lim_{k} x_{k} = \lim_{j} x_{n_{j}} .$

Both sides are asserted to exist: the right-hand side by hypothesis, the left-hand side as part of the conclusion.

Why this is a separate item. Each of the five is proved in this library for sequences of reals, and none of those proofs may be cited here. Conventions for sequences: indexing, eventually, $\lim$ , and rational $ε$ is explicit about it: a theorem about sequences of reals is a theorem about $R$ , and the fact that its argument would transfer to an arbitrary ordered field is a statement about the argument, not a licence to cite the result. The five are collected here, proved from the ordered field axioms alone, so that the completeness equivalences of this page have one place to cite instead of five inline reconstructions.

Facts & Assumptions

Given: An ordered field $F$ and sequences $(x_{k})$ , $(y_{k})$ in $F$ . Each of the five claims is proved under its own stated hypotheses; nothing is assumed of $(x_{k})$ or $(y_{k})$ outside the claim being proved.

[L1]

Sequences in an ordered field: $(x_{k})$ converges to $L$ in $F$ when for every $ε > 0$ in $F$ there is $N \in N$ with $∣ x_{k} - L ∣ < ε$ for all $k \geq N$ ; $(x_{k})$ is Cauchy in $F$ when for every $ε > 0$ in $F$ there is $N \in N$ with $∣ x_{k} - x_{l} ∣ < ε$ for all $k, l \geq N$ ; $(x_{k})$ is bounded when there is $M \in F$ with $∣ x_{k} ∣ \leq M$ for every $k$ ; and a subsequence of $(x_{k})$ is a sequence $(x_{n_{j}})_{j \in N}$ for a strictly increasing $n : N \to N$ (Sequences, convergence, Cauchyness, monotonicity, boundedness and closed intervals in an arbitrary ordered field).

[L2]

Triangle inequality: $∣ u + v ∣ \leq ∣ u ∣ + ∣ v ∣$ for $u, v \in F$ (The triangle inequality).

[L3]

Absolute value: $∣ u ∣ \geq 0$ ; $∣ u ∣ = 0$ if and only if $u = 0$ ; $∣ - u ∣ = ∣ u ∣$ ; and $u \leq ∣ u ∣$ (Basic properties of the absolute value).

[L4]

Order in $F$ : exactly one of $u < v$ , $u = v$ , $v < u$ holds, so the order is total, and both $<$ and $\leq$ are transitive; adding a constant preserves the strict order and two strict inequalities may be added (Order is preserved by adding a constant and by adding inequalities); the nonstrict forms of those two, used below, are the strict forms together with the equality cases, which trichotomy settles (Ordered field).

[L5]

Halving: $0 < 1_{F}$ (The multiplicative identity is positive), so $2 \cdot 1_{F} = 1_{F} + 1_{F} > 0$ (Canonical naturals are positive and strictly increasing) and $2 \cdot 1_{F}$ is nonzero, hence invertible with $(2 \cdot 1_{F})^{- 1} > 0$ (Inverses of positives are positive, and reciprocation reverses order). Writing $ε / 2$ for $ε \cdot (2 \cdot 1_{F})^{- 1}$ , an $ε > 0$ gives $ε / 2 > 0$ and $ε / 2 + ε / 2 = ε$ (Ordered field).

[L6]

Induction principle on $N$ (The principle of mathematical induction).

[L7]

Growth of an index map: a strictly increasing $n : N \to N$ satisfies $n_{j} \geq j$ for every $j$ (A strictly increasing index map satisfies $n_{k} \geq k$ ).

[L8]

The order on $N$ is total and transitive, so of any two indices one is $\geq$ the other, and every index $k$ satisfies $k \leq N$ or $k \geq N$ ( $\leq$ is a linear order on $N$ ).

Proof

technique · direct

1.1

If $d \in F$ satisfies $d < ε$ for every $ε > 0$ in $F$ , then $d \leq 0$ : were $d > 0$ , the instance $ε = d$ would give $d < d$ , which trichotomy forbids, so $d > 0$ fails and totality leaves $d \leq 0$ .

L4algebra

1.2

For every $ε > 0$ in $F$ one has $ε / 2 > 0$ and $ε / 2 + ε / 2 = ε$ .

1.3

Claim 1. Assume $x_{k} \to L$ and $x_{k} \to L^{'}$ , and let $ε > 0$ in $F$ be arbitrary; choose $N_{1}$ with $∣ x_{k} - L ∣ < ε / 2$ for $k \geq N_{1}$ , choose $N_{2}$ with $∣ x_{k} - L^{'} ∣ < ε / 2$ for $k \geq N_{2}$ , and let $N$ be whichever of $N_{1}, N_{2}$ is the larger.

L1L8choose

1.4

Claim 2. Assume $x_{k} \to L$ , $y_{k} \to M$ and $x_{k} \leq y_{k}$ for every $k$ , and let $ε > 0$ in $F$ be arbitrary; choose $N_{1}$ with $∣ x_{k} - L ∣ < ε / 2$ for $k \geq N_{1}$ , choose $N_{2}$ with $∣ y_{k} - M ∣ < ε / 2$ for $k \geq N_{2}$ , and let $N$ be the larger of the two.

L1L8choose

1.5

Claim 3. Assume $x_{k} \to L$ and let $ε > 0$ in $F$ be arbitrary; choose $N$ with $∣ x_{k} - L ∣ < ε / 2$ for all $k \geq N$ .

L1choose

1.6

Claim 4. For every $n \in N$ there is $B \in F$ with $∣ x_{j} ∣ \leq B$ for all $j \leq n$ , by induction on $n$ : for $n = 0$ take $B = ∣ x_{0} ∣$ ; and given such a $B$ for $n$ , totality of the order on $F$ gives either $∣ x_{n + 1} ∣ \leq B$ , in which case the same $B$ serves for $n + 1$ , or $B < ∣ x_{n + 1} ∣$ , in which case $∣ x_{n + 1} ∣$ serves for $n + 1$ by transitivity.

L1L4L6

1.7

Claim 4, continued. Assume $(x_{k})$ is Cauchy; since $1_{F} > 0$ , choose $N$ with $∣ x_{k} - x_{l} ∣ < 1_{F}$ for all $k, l \geq N$ , so that for $k \geq N$ one has $∣ x_{k} ∣ = ∣ (x_{k} - x_{N}) + x_{N} ∣ \leq ∣ x_{k} - x_{N} ∣ + ∣ x_{N} ∣ < 1_{F} + ∣ x_{N} ∣$ .

L1L2L4L5choose

1.8

Claim 5. Assume $(x_{k})$ is Cauchy and $x_{n_{j}} \to L$ along a strictly increasing $n$ , and let $ε > 0$ in $F$ be arbitrary; choose $N_{1}$ with $∣ x_{k} - x_{l} ∣ < ε / 2$ for $k, l \geq N_{1}$ , choose $N_{2}$ with $∣ x_{n_{j}} - L ∣ < ε / 2$ for $j \geq N_{2}$ , and let $N$ be the larger of the two, so that $n_{N} \geq N \geq N_{1}$ and $N \geq N_{2}$ .

L1L7L8choose

2.1

For every $k \geq N$ in the situation of step 1.3: $∣ L - L^{'} ∣ = ∣ (L - x_{k}) + (x_{k} - L^{'}) ∣ \leq ∣ L - x_{k} ∣ + ∣ x_{k} - L^{'} ∣ = ∣ x_{k} - L ∣ + ∣ x_{k} - L^{'} ∣ < ε / 2 + ε / 2 = ε$ .

step 1.2step 1.3L2L3L4

2.2

For every $k \geq N$ in the situation of step 1.4: $L - M = (L - x_{k}) + (x_{k} - y_{k}) + (y_{k} - M)$ , where $L - x_{k} \leq ∣ L - x_{k} ∣ < ε / 2$ and $y_{k} - M \leq ∣ y_{k} - M ∣ < ε / 2$ and $x_{k} - y_{k} \leq 0$ ; adding, $L - M < ε$ .

step 1.2step 1.4L3L4

2.3

For all $k, l \geq N$ in the situation of step 1.5: $∣ x_{k} - x_{l} ∣ = ∣ (x_{k} - L) + (L - x_{l}) ∣ \leq ∣ x_{k} - L ∣ + ∣ x_{l} - L ∣ < ε / 2 + ε / 2 = ε$ .

step 1.2step 1.5L2L3L4

2.4

In the situation of steps 1.6 and 1.7, let $B$ be a bound for $∣ x_{j} ∣$ over $j \leq N$ and set $M : = B + 1_{F} + ∣ x_{N} ∣$ ; then $B \geq ∣ x_{0} ∣ \geq 0$ and $1_{F} + ∣ x_{N} ∣ > 0$ , so $M \geq B$ and $M \geq 1_{F} + ∣ x_{N} ∣$ , whence $∣ x_{k} ∣ \leq B \leq M$ for $k \leq N$ and $∣ x_{k} ∣ < 1_{F} + ∣ x_{N} ∣ \leq M$ for $k \geq N$ ; as every index satisfies $k \leq N$ or $k \geq N$ , $(x_{k})$ is bounded.

step 1.6step 1.7L1L3L4L8

2.5

For every $k \geq N$ in the situation of step 1.8: $∣ x_{k} - L ∣ = ∣ (x_{k} - x_{n_{N}}) + (x_{n_{N}} - L) ∣ \leq ∣ x_{k} - x_{n_{N}} ∣ + ∣ x_{n_{N}} - L ∣ < ε / 2 + ε / 2 = ε$ , the first summand being covered because $k \geq N \geq N_{1}$ and $n_{N} \geq N_{1}$ .

step 1.2step 1.8L2L4

3.1

By step 2.1 the element $∣ L - L^{'} ∣$ is below every $ε > 0$ , so $∣ L - L^{'} ∣ \leq 0$ ; with $∣ L - L^{'} ∣ \geq 0$ this forces $∣ L - L^{'} ∣ = 0$ and hence $L = L^{'}$ , which is claim 1.

step 1.1step 2.1L3L4

3.2

By step 2.2 the element $L - M$ is below every $ε > 0$ , so $L - M \leq 0$ , that is $L \leq M$ , which is claim 2.

step 1.1step 2.2L4

3.3

Step 2.3 produced, for an arbitrary $ε > 0$ , an $N$ beyond which all pairs are within $ε$ , so $(x_{k})$ is Cauchy in $F$ , which is claim 3.

step 2.3L1

4.1

Step 2.5 produced, for an arbitrary $ε > 0$ , an $N$ beyond which $∣ x_{k} - L ∣ < ε$ , so $(x_{k})$ converges in $F$ with $x_{k} \to L$ ; since also $x_{n_{j}} \to L$ , step 3.1 identifies both limits as $L$ and gives $\lim_{k} x_{k} = \lim_{j} x_{n_{j}}$ , which is claim 5.

step 2.5step 3.1L1

5.1

Claims 1, 2, 3, 4 and 5 are steps 3.1, 3.2, 3.3, 2.4 and 4.1 respectively, so all five hold.

step 2.4step 3.1step 3.2step 3.3step 4.1∎

Remarks

Nothing above uses the Archimedean property, and nothing above uses completeness. The five claims hold in every ordered field, including $R (t)$ and $R ((t^{- 1}))$ . That is what makes them safe to use on both sides of every implication proved on this page.
Claim 2 is genuinely non-strict. From $x_{k} < y_{k}$ at every index one gets only $L \leq M$ : the sequences $x_{k} = 0$ and $y_{k} = ε / (k + 1)$ in an Archimedean $F$ have $x_{k} < y_{k}$ and equal limits. The real-number version of this warning is recorded at Limits preserve non-strict inequalities.
There is deliberately no arithmetic clause here. Nothing above lets one add, multiply or divide two limits in a general ordered field, and no item in this library does: Algebra of limits: sums, scalar multiples, products and quotients is stated for sequences of reals, and by the rule recalled above it may not be cited for a general $F$ . No proof on this page needs such a clause; every abstract argument here works with the defining $ε$ and $N$ directly, or with clauses 1 to 5.
Claim 4 avoids any appeal to a maximum of a finite set. The library's finite-maximum lemma Every nonempty finite set of reals has a maximum and a minimum is stated for $R$ , so it is unavailable here for the same reason the other four real-valued lemmas are; step 1.6 replaces it by an induction that uses nothing but totality of the order of $F$ .

Depends on

Used by

Dependency tree · two levels

60 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Ordered field (Wikipedia) (standard reference, not scraped)
Cauchy sequence (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 3 (standard reference, not scraped)
J. Lebl, Basic Analysis I, §2.1 and §2.4 (standard reference, not scraped)